Question

Let S1 : x² + y² = 9 and S2 : (x - 2)² + y² = 1. Then the locus of center of a variable circle S which touches S1 internally and S2 externally always passes through the points :

• A. (2, 3/2)
• B. (0, ±√3)
• C. (1, ±2)
• D. (1/2, ± √5 /2)

Hint:

The locus of the center of a circle touching two fixed circles is always a conic section (ellipse or hyperbola).

Answer 1

The Correct Option is B

The given problem involves two circles and requires us to find the locus of the center of a circle that touches these given circles under specific conditions.

Let's break down the problem:

1. Identify the circles:
   • The first circle S_1 is given by the equation x^2 + y^2 = 9. This circle has a center at (0, 0) and a radius of 3.
   • The second circle S_2 is represented by the equation (x - 2)^2 + y^2 = 1. Its center is at (2, 0) with a radius of 1.
2. Understanding the conditions for the variable circle:
   • The circle S touches S_1 internally. This implies that the distance between the center of circle S and the center of S_1 is the radius of S_1 minus the radius of S. Let r be the radius of circle S, and its center is (h, k). Then, \sqrt{h^2 + k^2} = 3 - r.
   • The circle S touches S_2 externally, meaning the distance between their centers is the sum of their radii: \sqrt{(h-2)^2 + k^2} = 1 + r.
3. Finding the locus:
   • Using the condition for internal tangency with S_1, we have: h^2 + k^2 = (3 - r)^2.
   • Using the condition for external tangency with S_2, we have: (h-2)^2 + k^2 = (1 + r)^2.
4. Combine and simplify the two equations:
   • From h^2 + k^2 = 9 - 6r + r^2 and (h-2)^2 + k^2 = 1 + 2r + r^2, subtract the second equation from the first:
   • \quad h^2 + k^2 - (h-2)^2 - k^2 = 9 - 6r + r^2 - 1 - 2r - r^2
   • \quad h^2 - (h^2 - 4h + 4) = 8 - 8r
   • \quad 4h - 4 = 8 - 8r
   • Solving gives: h = 2 - 2r.
   • Substituting back, we find h = 0, thus k^2 = 3.
5. Conclusion:
   • The points (0, \pm\sqrt{3}) satisfy the conditions, thereby confirming they are part of the locus of the center of S.

Therefore, the correct answer is the option (0, ±√3).